How to show:

In a acute angled $\triangle \ ABC$ show that $$\tan(A) \cdot \tan(B)\cdot \tan(C) \geq 3\sqrt{3}$$

Any ideas?

  • 4
    $\begingroup$ Use $\tan A \cdot \tan B \cdot \tan C = \tan A + \tan B + \tan C$, and the AM-GM inequality. $\endgroup$
    – Srivatsan
    Aug 13, 2011 at 6:45
  • 1
    $\begingroup$ The question math.stackexchange.com/questions/8732/… includes the identity Srivatsan mentions. $\endgroup$ Aug 13, 2011 at 6:56

3 Answers 3


These may be useful:

  • Since $\triangle ABC$ is acute, we have $\tan(A),\tan(B),\tan(C)$ positive.

  • By A.M-G.M you have $$\displaystyle \frac{\tan(A)+\tan(B)+\tan(C)}{3} \geq \sqrt[3]{\tan(A)\cdot\tan(B)\cdot\tan(C)}$$

  • $\text{The equality holds if the triangle is}$ $\textbf{equilateral.}$



\begin{align*} &\tan (A+B)= \tan (\pi-C)\\ &(\tan A+ \tan B)/(1-\tan A \tan B)= (\tan \pi- \tan C)/(1+\tan \pi \tan C)=-\tan C\\ &(\tan A+ \tan B)= -\tan C(1-\tan A \tan B)\\ &\tan A + \tan B= -\tan C+ \tan A \tan B \tan C\\ &\tan A + \tan B+ \tan C= \tan A \tan B \tan C \end{align*}

  • $\begingroup$ Now u can use the am-gm inequality as others have stated. and no problem in using that inequality as all the variables involved iwll be positive quantities in case of an acute angled traingle $\endgroup$
    – Bhargav
    Aug 13, 2011 at 6:52
  • $\begingroup$ You mean $A+B=\pi - C$, I guess. It would also be good if you indicated where exactly you need to use that $A,B,C \lt \pi/2$. $\endgroup$
    – t.b.
    Aug 13, 2011 at 6:54
  • 1
    $\begingroup$ $A+B+C=\pi$, not $2\pi$. Fortunately, $\tan(\pi)=\tan(2\pi)$. $\endgroup$ Aug 13, 2011 at 6:55


Use the AM-GM inequality.


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